Conservative forces on a sliding block

[peaceandlove]

Homework Statement

A 4.5 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 5.3 m/s, the height difference is h = 1.0 m, and μk = 0.613. Find d.

Homework Equations

E-mec 1 (object moving) = K1 + U1
E-mec 2 (object stopped) = K2 + U2 + energy lost to friction

The Attempt at a Solution

K1 + U1 - energy lost to friction = K2 + U2
I said that K1=(1/2)mv^2 and U1=mgy and the energy lost to friction is (fk)d; however, I don't know how to get the right side of the equation. I tried using the same formulas and I came up with 1.6313, but apparently that is incorrect.

 

[Hootenanny]

K1 + U1 - energy lost to friction = K2 + U2

This looks good. So you explicitly we have:

\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + R\mu_k d

And you want to find the distance, i.e. you want to solve this equation for d. What's the first thing you usually do when solving an equation?

 

[peaceandlove]

The first thing I would do is plug in all the values I know.

m=4.5kg
v1=5.3m/s
g=9.8m/s^2
h1=1m (not entirely sure about this one...)
v2=(I would think it to still be 5.3m/s, but I could be wrong)
h2=0 (I think...)
R=(no clue.)
Uk=0.613

 

[Hootenanny]

m=4.5kg
v1=5.3m/s
g=9.8m/s^2

Good.

h1=1m (not entirely sure about this one...)
h2=0 (I think...)

Don't worry about these, we'll get to these later.

v2=(I would think it to still be 5.3m/s, but I could be wrong)

Hint: You are looking for the distance d when the object has stopped.

R=(no clue.)

What is the expression for the maximum frictional force on an object?

Uk=0.613

Good.

Personally, I wouldn't plug in any of the values in just yet. Since we're solving for d I would try to make d the subject of the expression.

 

[peaceandlove]

So would v2 be 0 and R=(Us)(Fn)?

 

[peaceandlove]

And d=((1/2)m(v1)^2+mg(h1)-(1/2)m(v2)^2-mg(h2))/R(Uk).

 

[Hootenanny]

So would v2 be 0 and R=(Us)(Fn)?

Sounds good to me

And d=((1/2)m(v1)^2+mg(h1)-(1/2)m(v2)^2-mg(h2))/R(Uk).

Yup. However, you can write it in a somewhat nicer fashion:

\begin{align*}<br /> d &amp; = \frac{1}{R\mu_k}\left\{\frac{1}{2}mv_1^2 + mgh_1 - \frac{1}{2}mv_2^2 -mgh_2\right\} \\<br /> &amp; = \frac{1}{R\mu_k}\left\{\frac{1}{2}m\left(v_1^2 - v_2^2\right) + mg\left(h_1-h_2\right)\right\} \\<br /> &amp; = \frac{1}{R\mu_k}\left\{\frac{1}{2}m\left(v_1^2 - v_2^2\right) + mg\Delta h\right\}\end{align*}

Does that help?

 

[peaceandlove]

One last thing, what values do we plug in for (Us) and (Fn)? I got (Fn) to be 44.1 (although I could be wrong), but I don't know how to find (Us).

 

[Hootenanny]

One last thing, what values do we plug in for (Us) and (Fn)? I got (Fn) to be 44.1 (although I could be wrong), but I don't know how to find (Us).

Sorry, I misread your previous post. R is simply Fn= 44.1 N.

As a side note, take care with the sign of \Delta h, recall it's definition \Delta h = h_1-h_2.

 

[peaceandlove]

Good.

Don't worry about these, we'll get to these later.

Hint: You are looking for the distance d when the object has stopped.

What is the expression for the maximum frictional force on an object?

Good.

Personally, I wouldn't plug in any of the values in just yet. Since we're solving for d I would try to make d the subject of the expression.

Oh, and you never got back to clarifying what h1 and h2 are.

 

[Hootenanny]

Oh, and you never got back to clarifying what h1 and h2 are.

What do h₁ and h₂ represent?

 

[peaceandlove]

The change in y, right? So h1 is 1 and h2 is 0?

 

[Hootenanny]

The change in y, right? So h1 is 1 and h2 is 0?

Almost. h₁ and h₂ represent the initial and final heights respectively. Hence, \Delta h = h_1-h_2 represent the change in height. However, you should note that the question states that the final height is greater than the initial height.

 

[peaceandlove]

Oh, so it's the other way around? Meaning (delta)h=-1?

 

[Hootenanny]

Oh, so it's the other way around? Meaning (delta)h=-1?

Yes.

 

[peaceandlove]

Thank you so much for your help!

 

[Hootenanny]

Thank you so much for your help!

A pleasure